Page 72 - Applied statistics and probability for engineers

50 Chapter 2/Probability

so the probability of a surface law does not depend on whether the part is defective. Furthermore, the dei nition of
conditional probability implies that
P D F P F ( )
(
P F ∩ D) = ( | )
but in the special case of this problem,
(
P F ∩ D) = ( ) ( 2 . 2 = 1
P D P F) =
40 20 200
5"#-& t 2-4 Parts Classified

Surface Flaws
Yes (event F) No Total
Defective Yes (event D) 2 18 20
No 38 342 380
Total 40 360 400

The preceding example illustrates the following conclusions. In the special case that
| (
P B A) = ( )
P B , we obtain
(
P B A P A ( ) = ( ) (
P A∩ B) = ( | ) P B P A)
and P A∩ B) P A P B ( )
( )
(
| (
P A B) = = = P A ( )
P B ( ) P B ( )
These conclusions lead to an important dei nition.
Independence
(two events) Two events are independent if any one of the following equivalent statements is true:
(1) P A B| ( ) = P A ( )
(2) P B A| ( ) = P B ( )
(
(3) P A∩ B) = ( ) ( (2-13)
P A P B)

It is left as a mind-expanding exercise to show that independence implies related results such as
P A P B′ ( )
(
P A′ ∩ B′) = ( ′)
The concept of independence is an important relationship between events and is used
throughout this text. A mutually exclusive relationship between two events is based only on
the outcomes that compose the events. However, an independence relationship depends on the
probability model used for the random experiment. Often, independence is assumed to be part
of the random experiment that describes the physical system under study.

Example 2-31 Consider the inspection described in Example 2-14. Six parts are selected ran domly without replace-
ment from a bin of 50 parts. The bin contains 3 defective parts and 47 nondefective parts. Let A and
B denote the events that the irst and second parts are defective, respectively.

We suspect that these two events are not independent because the knowledge that the irst part is defective suggests that it

is less likely that the second part selected is defective. Indeed, P(B | A) = 2/49. Now, what is P(B)? Finding the unconditional

P(B) takes some work because the possible values of the irst selection need to be considered:
′
′
P B) = P B A P A) + P B A P A )
)
(
(
(
|
(
(
|
)
⋅
= 2 ⋅ 3 + 3 47 = 3
49 50 49 50 50